There is no "pull" of gravity, only the PUSH of flowing ether!
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From: Timo Nieminen on 7 May 2010 17:29 On May 8, 5:57 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 7, 2:21 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > Dear Timo: On the one hand you compliment me; on the other you chide > me for not having all of the numbers at my fingertips. No, I tell you that all of these numbers are available on www, so you don't even need to go and look in a book. When you sit at your keyboard, the relevant numbers _are_ at your fingertips. It's an obvious test. Since you claim your theory explains reality, I expected that you would have compared the two - your theory and reality - and checked if they agree. My mistake - you don't seem to have done this. You give a clear and easily checked statement: > Since gravity is > directly proportional to photon emission (not gravitons, which dont > exist), then it is the luminosity and the temperature of the light > that determine the gravity of stars. But before it's worth checking this, you should clarify: (1) By "directly proportional", you mean: gravity = (constant) times (photon emission)? That's is, linear proportionality. Or do you mean something else? (2) What do you mean by "photon emission"? What you mean by "photon" might not be what conventional science means by "photon". How is "photon emission" related to radiated power? Since the bolometric luminosity is the total radiated power, is there any _further_ dependence on temperature beyond its effect on the bolometric luminosity. (If talking about visual luminosity, then, yes, the bolometric luminosity depends on the visual luminosity and the temperature.) > At room temperatures gravity is mass proportional, and matches > Newtons law. There has to be an object-size threshold that DENIES > mass in favor of surface area and temperature. This is new. You didn't say anything about this that I saw before. > I suspect that a > heated Cavendish ball will have gravity somewhere between the room > temperature, and the white hot. "Suspect" isn't good enough for the experiment to be worthwhile. It's directly connected to the following point which you didn't address in your reply. Do note that this is absolutely essential for the experiment to be worthwhile (as you will no doubt already know, since this is a simple matter of logic and analytical ability). To repeat the question: if a Cavendish experiment _doesn't_ detect a greater gravitational force, what does that mean for your theory? > > > Consider this: If you can heat one ball white hot, > > > and you DO detect a greater gravity, youve confirmed my theory. > > > It would _support_ your theory, not confirm it in any absolute sense. > > If one tries this and _doesn't_ detect a greater gravitational force, > > would that mean your theory is wrong and it's time to forget it and move > > on?
From: PD on 7 May 2010 18:07 On May 7, 3:16 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 7, 9:12 am, PD <thedraperfam...(a)gmail.com> wrote: > > PD, you are a LIAR! Never ONCE have you explained why KE = 1/2mv^2 > isn't in violation of the Law of the Conservation of Energy. Until > you do (and you CAN'T) everyone will know that you are just an air- > head FRAUD! NoEinstein Oh, but I have. If you really need to have it explained again, I ask you this time to print it out. The law of conservation of energy says that any change in the energy of a system must be due solely to the work done on the system. The work is the force acting on the object times the displacement of the object. So any change in energy of the object must be due solely to this work. In the case of a falling body released from rest, we'll look at the increase in the kinetic energy, which must be due to the work done by the only force acting on the body -- gravity. If the increase of kinetic energy the body has at any time is accounted for by the work that was done on the body during that time, then we know that the law of conservation of energy has been respected. In the first second, the body will fall 16 ft. In the next second, it will fall an additional 48 feet. In the third second, it will fall an additional 80 feet. During these first three seconds, the force has remained constant, so that it is the same in the first second, the second second, the third second. The speed increases linearly, so that it is falling at 32 ft/s after the first second, 64 ft/s after the second second, and 96 ft/s after the third second. Now, let's take a look at the work. The work done since the drop, after the first second, is the force of gravity times the displacement. This is mass x g x (16 ft). So this is how much kinetic energy the object has after one second. Now, in the second second, we'll add more work, in the amount mass x g x (48 ft), since that's the displacement for the next second. This increases the kinetic energy of the body, so that it now has kinetic energy mass x g x (16 ft + 48 ft) = mass x g x (64 ft), and that number is four times bigger than it was after the first second. Now, in the third second, we'll add more work, in the amount mass x g x (80 ft), since that's the displacement for the next sentence. Since energy is conserved, this added energy must add to the kinetic energy of the body, so that it now has kinetic energy mass x g x (64 ft + 80 ft) = mass x g x (144 ft), and that number is nine times bigger than it was after the first second. Now, it should be plain that the kinetic energy is conserved, since the only thing that has been contributing to it is the work done in subsequent seconds. We lost nothing, and we added only that which gravity added. The energy is conserved. It should also be apparent that the kinetic energy is increasing in the ratios 1:4:9. Meanwhile, the velocities are increasing linearly, in the ratios 1:2:3. Now, any fourth grader can see that we've completely conserved energy, losing track of nothing, and yet the kinetic energy is increasing as the square of the velocity. 1:4:9 are the squares of 1:2:3. There is no violation of conservation of kinetic energy, and yet KE is proportional to v^2. Now, don't you feel silly that a 4th grader can understand all of this, but you've never understood it? > > > > > On May 6, 8:54 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 5, 11:43 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > OH? Then please explain, PD, how a UNIFORM force inputthe static > > > weight of the falling objectcan cause a semi-parabolic increase in > > > the KE. Haven't you heard?: Energy IN must = energy OUT! > > > NoEinstein > > > I have explained this to you dozens of times. I gather that you do not > > remember any of those posts, and you do not know how to use your > > newsreader or Google to go back and find any of those dozens of times > > when it has been explained to you. > > > I surmise that you are slipping into dementia, where each day begins > > anew, with any lessons learned the previous day forgotten. > > > I don't think it's a good use of my time to explain the same thing to > > you each day, only to have you retire at night and forget it by > > morning, do you? > > > PD > >
From: PD on 7 May 2010 18:13 On May 7, 3:02 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 7, 9:08 am, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: No. Since you are a fraud, I would be > happy if you could find, and paraphrase, even one bit of evidence > supporting, Lorentz. He and Einstein (ha!) were meant for each other! > NE Oh, this is easy. There is a circular track that circulates muons at a lab called g-2. Here is a picture of it, in case you doubt it's real: http://www.g-2.bnl.gov/pictures/g2magnet2.jpg The ring is about 30 feet across and about 90 feet around. Muons at rest live for 2.2 microseconds, which is easily observed with a Navy surplus oscilloscope. If the muons lived that long in the ring, they would go around the ring about 24 times before decaying. Instead, they go around 37 times. That is, they live longer when they are traveling fast around the ring. But the extra time they have before decaying is exactly what Lorentz time dilation says they will have. Perfect example of just one bit of evidence that time dilation is real. There is of course scads and scads of further evidence. There. Short and sweet, and indisputable. > > > > > On May 6, 8:42 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 5, 10:24 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear PD, the Parasite Dunce: If you have "other" supporting evidence > > > for Lorentz (ha!), paraphrase it! You are all bluster and no > > > substance! NoEinstein > > > John, as I said, there are SCORES of independent experiments that have > > all provided experimental evidence. You might as well be asking for a > > paraphrased summary of the support for Newton's laws of motion. > > > If you want to understand the depth of the experimental support, then > > you're going to have to immerse yourself in the OVERWHELMINGLY MASSIVE > > documentation of that support. That's the only way to truly convince > > yourself. > > > > > On May 4, 7:22 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 4, 11:38 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > Whatever the truth is, PD contorts it. "Rubber Rulers" has no > > > > > supporting experiment! Lorentz, the imbecile, used RR to 'explain' > > > > > the nil results of M-M. Then, supposed scientists say that M-M > > > > > SUPPORTS Lorentz! Where are the brains, and WHERE is the scientific > > > > > method! NE > > > > > Of course there are supporting experiments, John. You seem to be under > > > > the impression that the MMX was the only experiment ever done to test > > > > relativity and that the whole of relativity rests on this one > > > > experiment, so that if you somehow fault the MMX, then all of > > > > relativity falls. > > > > > Nothing could be further from the truth, John. Relativity has been > > > > tested in scores of experiments, all independent of each other. > > > > > > > On May 3, 9:43 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On May 3, 11:51 am, PD <thedraperfam...(a)gmail.com> wrote:> On May 1, 8:25 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > > On May 1, 11:00 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > Dear PD, the Parasite Dunce: You just said that "physics isn't > > > > > > > > > determined by logic". Of course, you would think that! That's > > > > > > > > > because you don't know HOW to reason! > > > > > > > > > Well, it's because physics is a science, which means that it invokes > > > > > > > > the scientific method, and it determines truth by experimental test, > > > > > > > > not by logic. > > > > > > > > Dear PD: WHERE was the "scientific method" when Lorentz proposed his > > > > > > > ANTI-ENGINEERING, "rubber ruler" explanation for the nil results of M- > > > > > > > M? > > > > > > > Lorentz's proposal was subject to experimental test, NoEinstein.. > > > > > > That's how science works. > > > > > > And what on earth makes you think that this stuff is "anti- > > > > > > engineering"? > > > > > > Perhaps you don't know that engineers make use of relativity in their > > > > > > designs whenever it is needed? If it's anti-engineering, why are > > > > > > engineers happy to use it as needed? > > > > > > > > And where was the scientific method when both Coriolis and > > > > > > > Einstein wrote energy equations that were exponential, and thus in > > > > > > > violation of the Law of the Conservation of Energy? > > > > > > > Those energy equations have also been thoroughly tested in experiment, > > > > > > John, exactly as I was stating. You on the other hand are trying to > > > > > > rule them out with your bandy-legged logic, rather than considering > > > > > > independently verified experimental tests. > > > > > > > > When the truth be > > > > > > > known, PD, is this low I. Q. flunky who compensates by constantly > > > > > > > faulting his superiors. He has never stated a single contribution > > > > > > > that he has made to science. For one who devotes so much time to... > > > > > > > 'science' shouldn't PD have... "something" to show for it? > > > > > > > NoEinstein > > > > > > > What do you think I should have to show for it, John? > > > > > > > PD- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > >
From: PD on 7 May 2010 18:15 On May 7, 3:35 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 7, 12:47 pm, PD <thedraperfam...(a)gmail.com> wrote: > > PD: Alright, then. What IS momentum? You have the floor to showcase > your stupidity. NE I've just explained that elsewhere in another post. Perhaps you can use your tools properly to find it. > > > > > On May 6, 9:23 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 5, 12:36 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > PD: The L. C. catalogue card number is: 5241857. (look on page 19). > > > Here's the response to my query at the Library of Congress: > > The LCCN you entered [ 5241857 ] was not found in the Library of > > Congress Online Catalog. > > Are you lying, John? > > What's the ISBN? > > > > Also, my The Wiley Engineer's Desk Reference, by Stanford I. Heisler, > > > on page 94, says momentum = mv. > > > That is different than F=mv. Momentum is not force. > > > Moreover, this is not a good definition of momentum, though it is a > > useful approximation for engineers, not suitable for physics. > > > > A scripted style of the "m" is used > > > to differentiate from "mass". That book errs by saying that the > > > "units" is: (mass)-feet/secondwhich is bullshit! > > > And yet you would have me trust this Wiley Engineer's Desk Reference, > > when you don't believe it yourself. When are you going to support any > > of your assertions, John, other than blustering about what comes out > > of your own head? > > > > Momentum is > > > measured in pounds! It is velocity proportional, and that is a > > > simple, unit-less FRACTION NE > > > > > On May 5, 2:56 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 4, 2:53 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > PD loves to extrapolate things into unworkability, so he can claim > > > > > everything was invalid. MOMENTUM is: F = mv, expressed in pounds. > > > > > He'll find that same equation (but not the correct units, pounds) in > > > > > most textbooks. NE > > > > > No, I won't, John. That equation F=mv is not listed in most > > > > textbooks. > > > > When you can clearly identify which title you think DOES have that > > > > listed, then I can look for myself. > > > > As it is, since you obviously have problems reading an understanding a > > > > single sentence from beginning to end, I have my doubts. > > > > > > > On May 4, 1:07 pm, af...(a)FreeNet.Carleton.CA (John Park) wrote: > > > > > > > > PD (thedraperfam...(a)gmail.com) writes: > > > > > > > > On May 3, 10:07=A0pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > >> Dear PD: =A0A thin "College Outline Series" book (that fits into the > > > > > > > >> bookcase behind my computer chair) entitled "Physics", by Clarence E > > > > > > > >> Bennett, states on page 19: "G. =A0Momentum and Impulse. =A0(1.) =A0Momen= > > > > > > > > tum > > > > > > > >> is defined as the product of the mass times velocity (mv)...." =A0The > > > > > > > >> letter F is used for momentum, because the equation defines forces. =A0= > > > > > > > > =97 > > > > > > > >> NoEinstein =97 > > > > > > > > > Oh, good grief. John, what is the ISBN on this book? I'd like to > > > > > > > > secure it to look at it. > > > > > > > > From what it is you just told me is in it, if I can verify that you > > > > > > > > can indeed read it correctly, it is a horrible, horrible booklet and > > > > > > > > should be burned as worthless. > > > > > > > > To quote the Spartans on a quite different occasion: If. > > > > > > > > I can't help noticing that the actual quoted passage is reasonable and > > > > > > > the inference about forces is purely in NE's words. > > > > > > > Exactly. > > > > > > > For what it's worth, momentum's *definition* is not mv, either. > > > > > > Electromagnetic fields have momentum, but this expression certainly > > > > > > does not work for them. The formula works for a certain class of > > > > > > matter-based objects traveling at low speed, and that's it. > > > > > > > PD- Hide quoted text - > > > > > > > - Show quoted text -- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > >
From: PD on 7 May 2010 18:16
On May 7, 3:35 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 7, 12:47 pm, PD <thedraperfam...(a)gmail.com> wrote: > > PD: Alright, then. What IS momentum? You have the floor to showcase > your stupidity. NE In the meantime, you could confess that what your reference actually says does not support in any way your ridiculous claim that F=mv. It's just something you made up. > > > > > On May 6, 9:23 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 5, 12:36 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > PD: The L. C. catalogue card number is: 5241857. (look on page 19). > > > Here's the response to my query at the Library of Congress: > > The LCCN you entered [ 5241857 ] was not found in the Library of > > Congress Online Catalog. > > Are you lying, John? > > What's the ISBN? > > > > Also, my The Wiley Engineer's Desk Reference, by Stanford I. Heisler, > > > on page 94, says momentum = mv. > > > That is different than F=mv. Momentum is not force. > > > Moreover, this is not a good definition of momentum, though it is a > > useful approximation for engineers, not suitable for physics. > > > > A scripted style of the "m" is used > > > to differentiate from "mass". That book errs by saying that the > > > "units" is: (mass)-feet/secondwhich is bullshit! > > > And yet you would have me trust this Wiley Engineer's Desk Reference, > > when you don't believe it yourself. When are you going to support any > > of your assertions, John, other than blustering about what comes out > > of your own head? > > > > Momentum is > > > measured in pounds! It is velocity proportional, and that is a > > > simple, unit-less FRACTION NE > > > > > On May 5, 2:56 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 4, 2:53 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > PD loves to extrapolate things into unworkability, so he can claim > > > > > everything was invalid. MOMENTUM is: F = mv, expressed in pounds. > > > > > He'll find that same equation (but not the correct units, pounds) in > > > > > most textbooks. NE > > > > > No, I won't, John. That equation F=mv is not listed in most > > > > textbooks. > > > > When you can clearly identify which title you think DOES have that > > > > listed, then I can look for myself. > > > > As it is, since you obviously have problems reading an understanding a > > > > single sentence from beginning to end, I have my doubts. > > > > > > > On May 4, 1:07 pm, af...(a)FreeNet.Carleton.CA (John Park) wrote: > > > > > > > > PD (thedraperfam...(a)gmail.com) writes: > > > > > > > > On May 3, 10:07=A0pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > >> Dear PD: =A0A thin "College Outline Series" book (that fits into the > > > > > > > >> bookcase behind my computer chair) entitled "Physics", by Clarence E > > > > > > > >> Bennett, states on page 19: "G. =A0Momentum and Impulse. =A0(1.) =A0Momen= > > > > > > > > tum > > > > > > > >> is defined as the product of the mass times velocity (mv)...." =A0The > > > > > > > >> letter F is used for momentum, because the equation defines forces. =A0= > > > > > > > > =97 > > > > > > > >> NoEinstein =97 > > > > > > > > > Oh, good grief. John, what is the ISBN on this book? I'd like to > > > > > > > > secure it to look at it. > > > > > > > > From what it is you just told me is in it, if I can verify that you > > > > > > > > can indeed read it correctly, it is a horrible, horrible booklet and > > > > > > > > should be burned as worthless. > > > > > > > > To quote the Spartans on a quite different occasion: If. > > > > > > > > I can't help noticing that the actual quoted passage is reasonable and > > > > > > > the inference about forces is purely in NE's words. > > > > > > > Exactly. > > > > > > > For what it's worth, momentum's *definition* is not mv, either. > > > > > > Electromagnetic fields have momentum, but this expression certainly > > > > > > does not work for them. The formula works for a certain class of > > > > > > matter-based objects traveling at low speed, and that's it. > > > > > > > PD- Hide quoted text - > > > > > > > - Show quoted text -- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > > |